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I needed help in trying to draw a poset such that $|\mu(x,y)| \gt 1$ for some pair of elements and, If we let $I$ be the linear poset on $2$ elements, how can we describe $\mu$ for the $k$-fold product poset $I^k$. And how to describe $\mu$ for the poset defined on the first $n$ numbers $\{1,\dots,n\}$ by the division relation.

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1 Answer 1

I’m assuming that you’re using this definition of $\mu(x,y)$:

$$\mu(x,y)=\begin{cases} 1,&\text{if }x=y\\ -\sum_{x\le z<y}\mu(x,z),&\text{if }x<y\\ 0,&\text{otherwise}\;. \end{cases}$$

Clearly, in order to get $|\mu(x,y)|>1$ you’re going to need to use the second clause of the definition. On the other hand, a simple chain won’t do the trick, so you need to ‘fatten up’ the partial order a bit. What if you try a four-element poset $\{a,b,c,d\}$ in which $a<b<d$ and $a<c<d$, but $b$ and $c$ are not comparable? Then $$\begin{align*} \mu(a,d)&=-\Big(\mu(a,a)+\mu(a,b)+\mu(a,c)\Big)\\ &=-\Big(1+(-1)+(-1)\Big)\\ &=1\;. \end{align*}\tag{1}$$

This isn’t quite good enough, but there’s a very simple generalization that does the trick.

For the poset $P_k=\{0,1\}^k$, I suggest first working out $\mu(0,x)$ for each $x\in P_k$, where $0$ is the minimum element $\langle 0,\dots,0\rangle$. You can do this in $k+1$ stages. First, $\mu(0,0)=1$, by definition. Now what is $\mu(0,x)$ if $x$ has exactly one non-zero coordinate? What if $\mu(0,x)$ if $x$ has exactly two non-zero coordinates? And so on. If you manage that successfully, you should be able to characterize $\mu(x,y)$ in terms of the places where $x$ and $y$ have non-zero coordinates.

For the third problem, where the ordering is by divisibility, $\mu(x,y)$ for $x\mid y$ is easily computed by looking at the prime factorization of the quotient $\frac{y}x$.

  • What if $\frac{y}x$ is prime? Then $\mu(x,y)=-\mu(x,x)=-1$.
  • What if $\frac{y}x=pq$ for distinct primes $p$ and $q$? Then $$\mu(x,y)=-\Big(\mu(x,x)+\mu(x,px)+\mu(x,qx)\Big)=1\;,$$ just as in $(1)$.
  • What if $\frac{y}x=p^2$ for some prime $p$? Then $\mu(x,y)=-\Big(\mu(x,x)+\mu(x,px)\Big)=0$.

Now try to generalize: what happens if $\frac{y}x$ has more than two prime factors? Build up one factor at a time, somewhat as I suggested building up one non-zero coordinate at a time in the previous problem. If you get completely stuck, look here, though you’ll still have to prove that the answer is right.

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